---
title: "Thevenin theorem, current in 10Ω"
subject: "Basic Electrical Engineering"
chapter: 3
key: "semester_1/basic_electrical_engineering:thevenin-current-10-ohm-20v-12v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/thevenin-current-10-ohm-20v-12v
site: BCT Engineering
---

# Thevenin theorem, current in 10Ω

Basic Electrical Engineering · Chapter 3: Network Theorems

- Asked in: 2071 Shrawan, 8 marks
- Video: [Thevenin's theorem Solved Example | Electric Circuits | Network Analysis | Network Theory](https://www.youtube.com/watch?v=tH9boIFwHW8) — Electrical and Electronics Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/thevenin-current-10-ohm-20v-12v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3

## Question

For the circuit shown in figure below, calculate the current in the 10 ohm resistance using Thevenin's theorem. [Figure: 5 Ω resistor in series with a 20 V source on the left; 2 Ω shunt in the middle; 8 Ω in series with a 12 V source on the top right; 10 Ω resistor on the right.]

## Short answer

Thevenin voltage is −6.286 V and Thevenin resistance is 9.429 Ω. The current in the 10 Ω resistor is 6.286/(9.429+10) ≈ 0.324 A, flowing upward.

## Answer

#### Reading the figure
Left branch: 5 Ω in series with the 20 V battery (+ on top) to the bottom rail. Middle: 2 Ω shunt from the top node N to the bottom rail. Right: 8 Ω in series with the 12 V battery (+ on the left, towards N), then down through the 10 Ω to the bottom rail. The 10 Ω is the load.

#### Step 1: Thevenin voltage (10 Ω removed)
No current in the 8 Ω or the 12 V battery, so

V_N = 20 × 2/(5 + 2) = 5.714 V (voltage divider of the 20 V source between 5 Ω and 2 Ω)

The 12 V battery has its + on the N side, so the open terminal is 12 V lower:

V_th = 5.714 − 12 = −6.286 V

The load terminal is 6.286 V below the bottom rail.

#### Step 2: Thevenin resistance (sources off)
R_th = 8 + (5 ∥ 2) = 8 + 1.4286 = 9.429 Ω

#### Step 3: Current in the 10 Ω
I = |V_th|/(R_th + 10) = 6.286/(9.429 + 10) = 6.286/19.429 = **0.3235 A**

The current flows upwards through the 10 Ω (from the bottom rail towards the 8 Ω side), because the open terminal is below the rail potential.

**Answer: I₁₀Ω ≈ 0.324 A**

Assumption: battery polarities are read from the plate lengths in the figure.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is V_N with the 10 ohm removed?
   A. 20 x 2/(5+2) = 5.714 V
   B. 20 V
   C. 12 V
   D. 8 V
   Answer: A. No current in the 8 ohm; divider of 20 V. (Key point: V_N)

2. How is V_th found?
   A. 5.714 + 12 = 17.7 V
   B. 5.714 - 12 = -6.286 V
   C. 12 V
   D. 5.714 V
   Answer: B. Open terminal is 12 V below N. (Key point: V_th)

3. What is R_th?
   A. 8 + 5 + 2 = 15 ohm
   B. 5||2 = 1.43 ohm
   C. 8 + (5 || 2) = 9.43 ohm
   D. 8 ohm
   Answer: C. Sources off. (Key point: R_th)

4. What is the 10 ohm current?
   A. 0.629 A
   B. 0.667 A
   C. 1.2 A
   D. 6.286/(9.429+10) = 0.324 A
   Answer: D. Divide |V_th| by total resistance. (Key point: Current)

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